Using the principle of mathematical induction, show that;
$2 + 4 + 6 + . . . . . + 2 n = n^{2} + n$

Open in App

Solution

Given; P(n) is 2 + 4 + 6 + …+ 2n = n² + n.
P(0) = 0 = 0²+ 0 ; it’s true.
P(1) = 2 = 1² + 1 ; it’s true.
P(2) = 2 + 4 = 2² + 2 ; it’s true.
P(3) = 2 + 4 + 6 = 3² + 2 ; it’s true.
Let P(k) be 2 + 4 + 6 + …+ 2k = k² + k is true;
⇒ P(k+1) is 2 + 4 + 6 + …+ 2k + 2(k+1) = k² + k + 2k +2
= (k² + 2k +1) + (k+1)
= $(k + 1)^{2} + (k + 1)$
⇒ P(k+1) is true when P(k) is true.
∴ By Mathematical Induction $2 + 4 + 6 + \dots + 2 n = n^{2} + n$ is true for all natural numbers n.

Suggest Corrections

Similar questions

2 + 4 + 6 + . . . . + 2 n = n^{2} + n

2 + 2^{2} + . . . . . + 2^{n} = 2^{n + 1} - 2

n .

Using the principle of mathematical induction, prove that

$1.2.3 + 2.3.4 + 3.4.5 + . . . + n (n + 1) (n + 2) = \frac{n (n + 1) (n + 2) (n + 3)}{4},$ for all $n \in N .$

1^{2} + 2^{2} + 3^{2} + . . . . + n^{2} > \frac{n^{2}}{3}

n \in N

Using principle of mathematical induction, prove that $41^{n} - 14^{n}$ is a multiple of 27.

Or
Prove by the principle of mathematical induction $n < 2^{n}$ for all $n ϵ N$ .

Join BYJU'S Learning Program